Type The Correct Answer In The Box. Use Numerals Instead Of Words.What Value Of $x$ Satisfies This Equation?$10(1.5)^{3x} = 80$Round Your Answer To The Nearest Hundredth.The Value Of $ X X X [/tex] Is $\square$.

by ADMIN 218 views

Understanding the Equation

To solve the given equation, we need to isolate the variable x. The equation is $10(1.5)^{3x} = 80$. Our goal is to find the value of x that satisfies this equation.

Isolating the Variable

The first step is to isolate the term with the variable x. We can do this by dividing both sides of the equation by 10, which gives us:

(1.5)3x=8(1.5)^{3x} = 8

Using Exponent Properties

Next, we can use the property of exponents that states $(am)n = a^{mn}$. In this case, we can rewrite the equation as:

1.53x=81.5^{3x} = 8

Taking the Logarithm

To solve for x, we can take the logarithm of both sides of the equation. We can use any base for the logarithm, but let's use the natural logarithm (ln) for this example:

ln(1.53x)=ln(8)\ln(1.5^{3x}) = \ln(8)

Applying the Logarithm Property

Using the property of logarithms that states $\ln(a^b) = b\ln(a)$, we can rewrite the equation as:

3xln(1.5)=ln(8)3x\ln(1.5) = \ln(8)

Solving for x

Now, we can solve for x by dividing both sides of the equation by 3 times the natural logarithm of 1.5:

x=ln(8)3ln(1.5)x = \frac{\ln(8)}{3\ln(1.5)}

Calculating the Value

To find the value of x, we can use a calculator to evaluate the expression:

x=ln(8)3ln(1.5)2.083×0.4052.081.2151.71x = \frac{\ln(8)}{3\ln(1.5)} \approx \frac{2.08}{3 \times 0.405} \approx \frac{2.08}{1.215} \approx 1.71

Rounding the Answer

Finally, we need to round the answer to the nearest hundredth. Therefore, the value of x is approximately 1.71.

The Final Answer

The value of x is 1.71\boxed{1.71}.

Conclusion

In this example, we used logarithms to solve an exponential equation. By isolating the variable x and using the properties of logarithms, we were able to find the value of x that satisfies the equation. This is a common technique used in mathematics to solve equations involving exponents and logarithms.

Tips and Tricks

  • When solving exponential equations, it's often helpful to use logarithms to isolate the variable.
  • Make sure to use the correct properties of logarithms, such as the power rule and the product rule.
  • When evaluating expressions involving logarithms, use a calculator to ensure accuracy.
  • Always round your answer to the nearest hundredth, unless specified otherwise.

Related Topics

  • Exponential equations
  • Logarithmic equations
  • Properties of logarithms
  • Solving equations involving exponents and logarithms

Further Reading

Frequently Asked Questions

Q: What is an exponential equation?

A: An exponential equation is an equation that involves an exponential expression, which is a number raised to a power. For example, the equation 2x=82^x = 8 is an exponential equation.

Q: How do I solve an exponential equation?

A: To solve an exponential equation, you can use logarithms to isolate the variable. This involves taking the logarithm of both sides of the equation and using the properties of logarithms to simplify the expression.

Q: What is the difference between an exponential equation and a logarithmic equation?

A: An exponential equation involves an exponential expression, while a logarithmic equation involves a logarithmic expression. For example, the equation 2x=82^x = 8 is an exponential equation, while the equation log2(x)=3\log_2(x) = 3 is a logarithmic equation.

Q: How do I use logarithms to solve an exponential equation?

A: To use logarithms to solve an exponential equation, you can take the logarithm of both sides of the equation and use the properties of logarithms to simplify the expression. For example, if you have the equation 2x=82^x = 8, you can take the logarithm of both sides and get log2(2x)=log2(8)\log_2(2^x) = \log_2(8).

Q: What are some common properties of logarithms that I should know?

A: Some common properties of logarithms include:

  • The power rule: loga(bc)=cloga(b)\log_a(b^c) = c\log_a(b)
  • The product rule: loga(bc)=loga(b)+loga(c)\log_a(bc) = \log_a(b) + \log_a(c)
  • The quotient rule: loga(bc)=loga(b)loga(c)\log_a(\frac{b}{c}) = \log_a(b) - \log_a(c)

Q: How do I use the properties of logarithms to simplify an expression?

A: To use the properties of logarithms to simplify an expression, you can apply the rules of logarithms to the expression. For example, if you have the expression log2(8x)\log_2(8x), you can use the product rule to simplify it to log2(8)+log2(x)\log_2(8) + \log_2(x).

Q: What is the difference between a natural logarithm and a common logarithm?

A: A natural logarithm is a logarithm with a base of ee, while a common logarithm is a logarithm with a base of 10. For example, the equation ln(x)=2\ln(x) = 2 is a natural logarithm, while the equation log10(x)=2\log_{10}(x) = 2 is a common logarithm.

Q: How do I convert between natural logarithms and common logarithms?

A: To convert between natural logarithms and common logarithms, you can use the following formulas:

  • ln(x)=log10(x)log10(e)\ln(x) = \frac{\log_{10}(x)}{\log_{10}(e)}
  • log10(x)=ln(x)ln(10)\log_{10}(x) = \frac{\ln(x)}{\ln(10)}

Q: What are some common mistakes to avoid when solving exponential equations?

A: Some common mistakes to avoid when solving exponential equations include:

  • Not using logarithms to isolate the variable
  • Not applying the properties of logarithms correctly
  • Not checking the domain of the logarithmic function
  • Not rounding the answer to the correct number of decimal places

Additional Resources

Conclusion

Solving exponential equations can be a challenging task, but with the right tools and techniques, it can be done. By using logarithms to isolate the variable and applying the properties of logarithms, you can simplify the expression and find the solution. Remember to check the domain of the logarithmic function and round the answer to the correct number of decimal places. With practice and patience, you can become proficient in solving exponential equations.